Question: The lifespans of tigers in a particular zoo are normally distributed. The average tiger lives $17.8$ years; the standard deviation is $3.6$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a tiger living between $7$ and $25$ years.
Explanation: $17.8$ $14.2$ $21.4$ $10.6$ $25$ $7$ $28.6$ $99.7\%$ $95\%$ $2.35\%$ $2.35\%$ We know the lifespans are normally distributed with an average lifespan of $17.8$ years. We know the standard deviation is $3.6$ years, so one standard deviation below the mean is $14.2$ years and one standard deviation above the mean is $21.4$ years. Two standard deviations below the mean is $10.6$ years and two standard deviations above the mean is $25$ years. Three standard deviations below the mean is $7$ years and three standard deviations above the mean is $28.6$ years. We are interested in the probability of a tiger living between $7$ and $25$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the tigers will have lifespans within 3 standard deviations of the average lifespan. It also tells us that $95\%$ of the tigers will have lifespans within 2 standard deviations of the mean. That leaves $99.7\% - 95\% = 4.7\%$ of tigers between 2 and 3 standard deviations of the mean, or $2.35\%$ on either side of the distribution. The probability of a particular tiger living between $7$ and $25$ years is $\color{orange}{2.35\%} + {95\%}$, or $97.35\%$.